Susan Sound predicts that students will learn most effectively with a constant b
ID: 3316159 • Letter: S
Question
Susan Sound predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound or no sound at all. She randomly divides twenty-four students into three groups of eight. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with noise that changes volume periodically. Those in group 3 study with no sound at all. After studying, all students take a 10 point multiple choice test over the material. Their scores follow: 2. test scores 1) constant sound 2) random sound 3) no sound a. Identify the followine (4 i. Experimental Unit i. Response Variable ili. Factor v. Levels of the Factor b. State the null and altemative hypotheses 14) HO: HA: c State the critical value, and level of significance lAssume a05) (2)Explanation / Answer
a.
Experimental Unit - Students
Response variable - Test Scores
Factor - Sound
Levels of factor - constant sound, random sound, no sound
b.
Null hypotheses H0 : The average test score of students is statistically equal across the three levels of sounds.
Alternative hypotheses H1 : Atleast one average test score of students is statistically different across the three levels of sounds.
c.
Level of significance = 0.05
Degree of freedom of group = Number of level - 1 = 3 - 1 = 2
Degree of freedom of error = Number of observations - Number of level = 24 - 3 = 21
Critical value of F at DF = 2,21 is 3.47
d.
Let Ti be the total test score for group i, ni be number of observations of group i.
Let G be the total test scores of all observations and N be total number of observations.
X2 for each treatment and the entire data set
T1 = 48, T2 = 32 , T3 = 27
G = 48 + 32 + 27 = 107
X2 = 322 + 148 + 125 = 595
SST = X2 - G2/N = 595 - 1072/24 = 117.96
SSTR = T2/n - G2/N = (482 /8 + 322 /8 + 272 /8 ) - 1072/24 = 30.08
SSE = 117.96 - 30.08 = 87.88
e.
As, p-value is less than the significance level (or the observed value of F (3.6) is greater than the critical value (3.47), we reject the null hypothesis and conclude that the atleast one average test score of students is statistically different across the three levels of sounds.
Source DF SS MS F Pr(>F) Group 2 30.08 15.04 3.6 0.0453 Error 21 87.88 4.18 Total 23 117.96Related Questions
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